SOLUTION: Every sack of sugar has the same weight. Every sack of flour has the same weight, but not necessarily the same as the weight of the sacks of sugar. Suppose that three sacks of suga
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Question 1076461: Every sack of sugar has the same weight. Every sack of flour has the same weight, but not necessarily the same as the weight of the sacks of sugar. Suppose that three sacks of sugar together with four sacks of flour weighs no more than 50 pounds, and that the weight of two sacks of flour is no more than 13 pounds more than the weight of three sacks of sugar. What is the largest possible weight (in pounds) of a sack of flour?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
your 2 inequalities are:
3x + 4y <= 50
2y <= 3x + 13
the second equation can be shown as:
-3x + 2y <= 13
solve the equality portion of these 2 inequalities.
those are the following equations:
3x + 4y = 50
-3x + 2y = 13
add these equations together and you get:
6y = 63
solve for y to get y = 10.5
that appears to be a break even point.
that value of y has to satisfy both inequalities.
when y = 10.5, the first inequality is solved as follows:
3x + 4y <= 50
replace y with 10.5 and the inequality becomes:
3x + 42 <= 50
solve for x to get x <= 2.67
when y = 10.5, the second inequality is solved as follows:
-3x + 2y <= 13
replace y with 10.5 and the inequality becomes:
-3x + 21 <= 13
solve for x to get x >= 2.67
this indicates that, when y = 10.5, x has to be equal to 2.67.
both constraints are satisfied.
it remains to be tested for y > 10.5 and for y < 10.5
you can pick any values of y < 10.5 and any values of y > 10.5
i picked y = 5 and y = 20
when y = 5:
3x + 4y <= 50 tells you that x has to be <= 10.
and:
-3x + 2y <= 13 tells you that x has to be >= -1.
since x can't be negative, then x has to be >= 0.
both inequalities are satisfied when y = 5 which is less than y = 10.5
when y = 20:
3x + 4y <= 50 tells you that x has to be <= -10.
since this is impossible, then you can assume that y can't be >= 10.5
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